Let $X$ be a Banach space and let $T: X \rightarrow X$ be a bounded linear map.

(a) Define the spectrum $\sigma(T)$, the resolvent set $\rho(T)$ and the point spectrum $\sigma_{p}(T)$ of $T$.

(b) What does it mean for $T$ to be a compact operator?

(c) Show that if $T$ is a compact operator on $X$ and $a>0$, then $T$ has at most finitely many linearly independent eigenvectors with eigenvalues having modulus larger than $a$.

[You may use without proof the fact that for any finite dimensional proper subspace $Y$ of a Banach space $Z$, there exists $x \in Z$ with $\|x\|=1$ and $\operatorname{dist}(x, Y)=\inf _{y \in Y}\|x-y\|=1$.]

(d) For a sequence $\left(\lambda_{n}\right)_{n \geqslant 1}$ of complex numbers, let $T: \ell^{2} \rightarrow \ell^{2}$ be defined by

$$T\left(x_{1}, x_{2}, \ldots\right)=\left(\lambda_{1} x_{1}, \lambda_{2} x_{2}, \ldots\right) .$$

Give necessary and sufficient conditions on the sequence $\left(\lambda_{n}\right)_{n} \geqslant 1$ for $T$ to be compact, and prove your assertion.